Integrand size = 22, antiderivative size = 23 \[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\frac {3 \left (-1+x^2\right ) \sqrt [3]{-x+x^3}}{2 x} \]
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Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1604} \[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\frac {3 \left (x^3-x\right )^{4/3}}{2 x^2} \]
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Rule 1604
Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-x+x^3\right )^{4/3}}{2 x^2} \\ \end{align*}
Time = 0.54 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\frac {3 \left (x \left (-1+x^2\right )\right )^{4/3}}{2 x^2} \]
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Time = 0.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
trager | \(\frac {3 \left (x^{2}-1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{2 x}\) | \(20\) |
pseudoelliptic | \(\frac {3 \left (x^{2}-1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{2 x}\) | \(20\) |
gosper | \(\frac {3 \left (1+x \right ) \left (x -1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{2 x}\) | \(21\) |
risch | \(\frac {3 {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{3}} \left (x^{4}-2 x^{2}+1\right )}{2 x \left (x^{2}-1\right )}\) | \(32\) |
meijerg | \(-\frac {3 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], x^{2}\right )}{2 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {2}{3}}}+\frac {9 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{2}\right )}{4 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}}}\) | \(66\) |
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\frac {3 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}}{2 \, x} \]
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\[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\int \frac {\sqrt [3]{x \left (x - 1\right ) \left (x + 1\right )} \left (3 x^{2} + 1\right )}{x^{2}}\, dx \]
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\[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\int { \frac {{\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (3 \, x^{2} + 1\right )}}{x^{2}} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\frac {3}{2} \, x^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - \frac {3}{2} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \]
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Time = 5.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (1+3 x^2\right ) \sqrt [3]{-x+x^3}}{x^2} \, dx=\frac {3\,{\left (x^3-x\right )}^{1/3}\,\left (x^2-1\right )}{2\,x} \]
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